Formatted question description: https://leetcode.ca/all/971.html

971. Flip Binary Tree To Match Preorder Traversal (Medium)

Given a binary tree with N nodes, each node has a different value from {1, ..., N}.

A node in this binary tree can be flipped by swapping the left child and the right child of that node.

Consider the sequence of N values reported by a preorder traversal starting from the root. Call such a sequence of N values the voyage of the tree.

(Recall that a preorder traversal of a node means we report the current node's value, then preorder-traverse the left child, then preorder-traverse the right child.)

Our goal is to flip the least number of nodes in the tree so that the voyage of the tree matches the voyage we are given.

If we can do so, then return a list of the values of all nodes flipped. You may return the answer in any order.

If we cannot do so, then return the list [-1].

Example 1:

Input: root = [1,2], voyage = [2,1]
Output: [-1]

Example 2:

Input: root = [1,2,3], voyage = [1,3,2]
Output: [1]

Example 3:

Input: root = [1,2,3], voyage = [1,2,3]
Output: []

Note:

1 <= N <= 100

Related Topics:
Tree, Depth-first Search

Solution 1.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> flipMatchVoyage(TreeNode root, int[] voyage) {
        if (root == null)
            return new ArrayList<Integer>();
        Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
        int length = voyage.length;
        for (int i = 0; i < length; i++)
            indexMap.put(voyage[i], i);
        List<Integer> flipList = new ArrayList<Integer>();
        int voyageIndex = 0;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while (!stack.isEmpty()) {
            if (voyageIndex >= length)
                return new ArrayList<Integer>(Arrays.asList(-1));
            TreeNode node = stack.pop();
            if (node.val != voyage[voyageIndex])
                return new ArrayList<Integer>(Arrays.asList(-1));
            voyageIndex++;
            TreeNode left = node.left, right = node.right;
            if (left != null && right != null && right.val == voyage[voyageIndex]) {
                flipList.add(node.val);
                stack.push(left);
                stack.push(right);
            } else {
                if (right != null)
                    stack.push(right);
                if (left != null)
                    stack.push(left);
            }
        }
        if (voyageIndex != length)
            return new ArrayList<Integer>(Arrays.asList(-1));
        return flipList;
    }
}

// OJ: https://leetcode.com/problems/flip-binary-tree-to-match-preorder-traversal/
// Time: O(NlogN)
// Space: O(logN)
class Solution {
private:
    vector<int> ans;
    int find(vector<int>& v, int begin, int end, int target) {
        int i = begin;
        while (i < end && v[i] != target) ++i;
        return i;
    }
    bool preorder(TreeNode *root, vector<int>& voyage, int begin, int end) {
        if (!root) return end == begin;
        if (end == begin || voyage[begin] != root->val) return false;
        if (!root->left && !root->right) return begin + 1 == end;
        int leftIndex = end, rightIndex = end;
        if (root->right) {
            rightIndex = find(voyage, begin + 1, end, root->right->val);
            if (rightIndex == end) return false;
        }
        if (root->left) {
            leftIndex = find(voyage, begin + 1, end, root->left->val);
            if (leftIndex == end) return false;
        }
        if (preorder(root->left, voyage, begin + 1, rightIndex)
                && preorder(root->right, voyage, rightIndex, end)) return true;
        if (preorder(root->left, voyage, leftIndex, end)
                && preorder(root->right, voyage, begin + 1, leftIndex)) {
            ans.push_back(root->val);
            return true;
        }
        return false;
    }
public:
    vector<int> flipMatchVoyage(TreeNode* root, vector<int>& voyage) {
        if (preorder(root, voyage, 0, voyage.size())) return ans;
        return { -1 };
    }
};

# 971. Flip Binary Tree To Match Preorder Traversal
# https://leetcode.com/problems/flip-binary-tree-to-match-preorder-traversal/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flipMatchVoyage(self, root: Optional[TreeNode], voyage: List[int]) -> List[int]:
        index = 0
        res = []
        
        def go(node):
            nonlocal index, res
            
            if not node: return True
            
            if node.val != voyage[index]: return False
            
            index += 1
            
            if node.left and node.right and node.left.val != voyage[index] and node.right.val == voyage[index]:
                res.append(node.val)
                
                node.left, node.right = node.right, node.left
            
            return go(node.left) and go(node.right)
        
        return res if go(root) else [-1]
            
            

971 - Flip Binary Tree To Match Preorder Traversal

971. Flip Binary Tree To Match Preorder Traversal (Medium)

Solution 1.

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971. Flip Binary Tree To Match Preorder Traversal (Medium)

Solution 1.

All Problems

All Solutions